Evaluate: `int(dx)/(xsqrt(x^(6)-16))=`

Options

A. `sec.(x^(3))/(4)+c`
B. `(1)/(12)sec.(x^(3))/(4)+c`
C. `(1)/(3)sec.(x^(3))/(4)+c`
D. none of these

Explanation

To evaluate the integral \( I = \int \frac{dx}{x \sqrt{x^6 - 16}} \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{x \sqrt{x^6 - 16}} \]
To simplify the integral, we can multiply and divide by \( x^2 \):
\[ I = \int \frac{x^2 \, dx}{x^3 \sqrt{x^6 - 16}} = \int \frac{x^2 \, dx}{x^3 \sqrt{x^6 - 16}} \]

### Step 2: Substitution
Let \( t = x^3 \). Then, differentiating both sides gives:
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
Substituting \( x^2 \, dx \) in terms of \( t \):
\[ x^2 \, dx = \frac{1}{3} dt \]
Now, substitute \( x^3 = t \) into the integral:
\[ I = \int \frac{\frac{1}{3} dt}{t \sqrt{t^2 - 16}} \]

### Step 3: Simplify the Integral
Now we can factor out the constant:
\[ I = \frac{1}{3} \int \frac{dt}{t \sqrt{t^2 - 16}} \]

### Step 4: Evaluate the Integral
The integral \( \int \frac{dt}{t \sqrt{t^2 - 16}} \) can be evaluated using the formula:
\[ \int \frac{dt}{t \sqrt{t^2 - a^2}} = \frac{1}{a} \sec^{-1} \left( \frac{t}{a} \right) + C \]
Here, \( a = 4 \):
\[ I = \frac{1}{3} \cdot \frac{1}{4} \sec^{-1} \left( \frac{t}{4} \right) + C = \frac{1}{12} \sec^{-1} \left( \frac{t}{4} \right) + C \]

### Step 5: Substitute Back
Since \( t = x^3 \), we substitute back:
\[ I = \frac{1}{12} \sec^{-1} \left( \frac{x^3}{4} \right) + C \]

### Final Answer
Thus, the evaluated integral is:
\[ I = \frac{1}{12} \sec^{-1} \left( \frac{x^3}{4} \right) + C \]